Question

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

If the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mg) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?

Answer 1

Answer:
`2(10)^(-9) mg`

Step-by-step explanation:

We know the total diameter of the cell (assumed spherical) is:

`d=1.9\mu m=1.9(10)^(-6) m`

Then its total radius
`r=(d)/(2)=(1.9(10)^(-6) m)/(2)=9.5(10)^(-7) m`

On the other hand, we know the thickness of the cell wall is
`r_(t)=60 nm= 60(10)^(-9) m` and its density is the same as water (
`\rho=997 kg/m^(3)`).

Since density is the relation between the mass
`m` and the volume
`V`:

`\rho=(m)/(V)`

The mass is:
`m=\rho V` (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

`V=(4)/(3)\pi R^(3)` (2)

Where
`R=r-r_(w)=9.5(10)^(-7) m - 60(10)^(-9) m`

Then:

`V=(4)/(3)\pi (9.5(10)^(-7) m - 60(10)^(-9) m)^(3)` (3)

`V=2.952(10)^(-18) m^(3)` (4)

Substituting (4) in (1):

`m=(997 kg/m^(3))(2.952(10)^(-18) m^(3))` (5)

`m=2.94(10)^(-15) kg` (6)

Knowing
`1 kg=1000 g` and
`1 mg=0.001 g`:

`m=2.94(10)^(-15) kg=2(10)^(-9) mg`